Integrability of the derivative of $f(x) = x^2 \sin (1/x^2)$ if $x \ne 0$ and 0 otherwise.

Question

why is the derivative of the function $f(x) = x^2 \sin (1/x^2)$ if $x \ne 0$ and 0 otherwise, not integrable?

Answer 1

For $x \ne 0$ we have $f'(x)=2x \sin(1/x^2)-\frac{1}{x}\cos(1/x^2)$. Then:

$f'(\frac{1}{\sqrt{n \pi}})=-\sqrt{n \pi}\cos(n \pi)=(-1)^{n+1}\sqrt{n \pi}$,

thus $|f'(\frac{1}{\sqrt{n \pi}})|=\sqrt{n \pi} \to \infty$ as $n \to \infty.$

Similar: $|f'(-\frac{1}{\sqrt{n \pi}})|=\sqrt{n \pi} \to \infty$ as $n \to \infty.$

Conclusion: if $[a,b]$ is an intervall with $0 \in [a,b]$ , then $f'$ is not bounded on $[a,b]$, hence $f'$ is not R- integrable on $[a,b]$.